My uni's complex analysis lecture notes simply state that the coefficients in a Laurent expansion are unique, and don't give a proof. It occurred to me to prove in the manner shown in this answer: Coefficients for Laurent series expansion are unique
However, it's not immediately clear to me at all why the Laurent coefficients of the 0 function are identically 0, and the answer given in that link doesn't give an explanation.
In short: why, if $\sum_{n=-\infty}^{n=\infty} a_nz^n=0$ for all $z$ in some annulus, do we have $a_n=0$ for all n?
For Taylor series there's the usual trick of differentiating $k$ times and evaluating at 0 to show that the $k$-th coefficient is 0, but of course that doesn't work here
Best Answer
The following proof is essentially taken from Wikipedia: Laurent series:
Let $f(z) = \sum_{n=-\infty}^\infty a_n z^n $ be a Laurent series in the annulus $r < |z| < R$ and assume that $f(z) = 0$ for all $z$.
Let $k$ be an integer and choose any $\rho \in (r, R)$. The series converges uniformly on $|z| = \rho$ so that we can change the order of integration and summation in the following calculation: $$ 0 = \frac{1}{2\pi i} \int_{|z| = \rho }\sum_{n=-\infty}^\infty a_n z^{n-k-1} \, dz = \sum_{n=-\infty}^\infty a_n \frac{1}{2\pi i} \int_{|z| = \rho } z^{n-k-1} \, dz = \sum_{n=-\infty}^\infty a_n \delta_{nk} = a_k \, , $$ i.e. all $a_k$ are zero.